Question

5v + 10 ≤ -4v - 17 < 9 - 2v
rewrite the compound inequality as two single inequalities and solve each one.
1.
2.

1. graph the solution set of the compound inequality.

kali just started a new sales floor job to save for college. she earns $15.75 per hour plus a flat fee of $50 per week. she wants to earn between $200 and $400 per week. the following inequality represents her earning potential:
200 ≤ 15.75x + 50 ≤ 400

1. solve the inequality.
2. graph the solution on the number line. approximate values as necessary.
3. complete the statement: kali can work between __ and __ hours each week.

Explanation:

Step1: Rewrite first compound - inequality

Rewrite $5v + 10\leq - 4v-17<9 - 2v$ as two inequalities:
$5v + 10\leq - 4v-17$ and $-4v - 17<9 - 2v$.
Solve $5v + 10\leq - 4v-17$:
Add $4v$ to both sides: $5v+4v + 10\leq-17$, which simplifies to $9v+10\leq - 17$.
Subtract 10 from both sides: $9v\leq-17 - 10$, so $9v\leq-27$.
Divide both sides by 9: $v\leq - 3$.
Solve $-4v - 17<9 - 2v$:
Add $4v$ to both sides: $-17<9 - 2v+4v$, which simplifies to $-17<9 + 2v$.
Subtract 9 from both sides: $-17 - 9<2v$, so $-26<2v$.
Divide both sides by 2: $v>-13$.
The solution for the first compound - inequality is $-13 < v\leq - 3$.

Step2: Solve second inequality

Rewrite $200\leq15.75x + 50\leq400$ as two inequalities:
$200\leq15.75x + 50$ and $15.75x + 50\leq400$.
Solve $200\leq15.75x + 50$:
Subtract 50 from both sides: $200 - 50\leq15.75x$, so $150\leq15.75x$.
Divide both sides by 15.75: $x\geq\frac{150}{15.75}=\frac{150}{\frac{63}{4}}=\frac{150\times4}{63}=\frac{200}{21}\approx9.52$.
Solve $15.75x + 50\leq400$:
Subtract 50 from both sides: $15.75x\leq400 - 50$, so $15.75x\leq350$.
Divide both sides by 15.75: $x\leq\frac{350}{15.75}=\frac{350}{\frac{63}{4}}=\frac{350\times4}{63}=\frac{200}{9}\approx22.22$.
The solution for the second inequality is $\frac{200}{21}\leq x\leq\frac{200}{9}$.

Step3: Graph first solution

On the number - line, for $-13 < v\leq - 3$, we have an open circle at $v=-13$ (because $v > - 13$) and a closed circle at $v = - 3$ (because $v\leq - 3$), and we shade the region between them.

Step4: Graph second solution

On the number - line, for $\frac{200}{21}\approx9.52\leq x\leq\frac{200}{9}\approx22.22$, we have a closed circle at $x=\frac{200}{21}$ and a closed circle at $x=\frac{200}{9}$, and we shade the region between them.

Step5: Complete the statement for Kali

Kali can work between $\frac{200}{21}\approx9.52$ and $\frac{200}{9}\approx22.22$ hours each week.

Answer:

1. Solution for $5v + 10\leq - 4v-17$ is $v\leq - 3$, solution for $-4v - 17<9 - 2v$ is $v>-13$, combined solution is $-13 < v\leq - 3$.
2. Solution for $200\leq15.75x + 50$ is $x\geq\frac{200}{21}\approx9.52$, solution for $15.75x + 50\leq400$ is $x\leq\frac{200}{9}\approx22.22$, combined solution is $\frac{200}{21}\leq x\leq\frac{200}{9}$.
3. For $-13 < v\leq - 3$, open circle at $-13$, closed circle at $-3$, shade between. For $\frac{200}{21}\leq x\leq\frac{200}{9}$, closed circles at both endpoints, shade between.
4. $\frac{200}{21}\leq x\leq\frac{200}{9}$
5. Closed circles at $\frac{200}{21}$ and $\frac{200}{9}$, shade between.
6. $\frac{200}{21}$; $\frac{200}{9}$